Example 1.  Solve the linear system of equations  
        [Graphics:Images/EchelonFormMod_gr_13.gif]    

Solution 1.

Enter the equations into Mathematica.  

[Graphics:../Images/EchelonFormMod_gr_14.gif]

Identify the matrix of coefficients A and column vector B for the matrix problem AX = B.  

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[Graphics:../Images/EchelonFormMod_gr_16.gif]

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Form the augmented matrix  M = [A, B]  and perform Gauss-Jordan elimination with row interchanges.

[Graphics:../Images/EchelonFormMod_gr_22.gif]



[Graphics:../Images/EchelonFormMod_gr_23.gif]

[Graphics:../Images/EchelonFormMod_gr_24.gif]

[Graphics:../Images/EchelonFormMod_gr_25.gif]

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[Graphics:../Images/EchelonFormMod_gr_27.gif]


[Graphics:../Images/EchelonFormMod_gr_28.gif]

[Graphics:../Images/EchelonFormMod_gr_29.gif]

[Graphics:../Images/EchelonFormMod_gr_30.gif]

[Graphics:../Images/EchelonFormMod_gr_31.gif]

Find the reduced row echelon form of the augmented matrix  M = [A, B].  

[Graphics:../Images/EchelonFormMod_gr_32.gif]


[Graphics:../Images/EchelonFormMod_gr_33.gif]

[Graphics:../Images/EchelonFormMod_gr_34.gif]


[Graphics:../Images/EchelonFormMod_gr_35.gif]

[Graphics:../Images/EchelonFormMod_gr_36.gif]

[Graphics:../Images/EchelonFormMod_gr_37.gif]

[Graphics:../Images/EchelonFormMod_gr_38.gif]

[Graphics:../Images/EchelonFormMod_gr_39.gif]

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The 3×3 identity matrix appears in the left 3 columns of  M, and the given linear system is equivalent to:

    [Graphics:../Images/EchelonFormMod_gr_43.gif]

The solution vector is the fourth column of  M.

[Graphics:../Images/EchelonFormMod_gr_44.gif]


[Graphics:../Images/EchelonFormMod_gr_45.gif]

Verify the solution.

[Graphics:../Images/EchelonFormMod_gr_46.gif]



[Graphics:../Images/EchelonFormMod_gr_47.gif]

[Graphics:../Images/EchelonFormMod_gr_48.gif]

[Graphics:../Images/EchelonFormMod_gr_49.gif]

[Graphics:../Images/EchelonFormMod_gr_50.gif]

[Graphics:../Images/EchelonFormMod_gr_51.gif]

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We are done.

Aside.  We can let Mathematica find the reduced row echelon matrix.  This is just for fun !

[Graphics:../Images/EchelonFormMod_gr_53.gif]


[Graphics:../Images/EchelonFormMod_gr_54.gif]

[Graphics:../Images/EchelonFormMod_gr_55.gif]

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Aside.  We check out  [Graphics:../Images/EchelonFormMod_gr_57.gif].  

[Graphics:../Images/EchelonFormMod_gr_58.gif]


[Graphics:../Images/EchelonFormMod_gr_59.gif]

[Graphics:../Images/EchelonFormMod_gr_60.gif]

[Graphics:../Images/EchelonFormMod_gr_61.gif]

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[Graphics:../Images/EchelonFormMod_gr_63.gif]

[Graphics:../Images/EchelonFormMod_gr_64.gif]

Looking at the above calculations we see that  [Graphics:../Images/EchelonFormMod_gr_65.gif]  and  [Graphics:../Images/EchelonFormMod_gr_66.gif].    
Since  [Graphics:../Images/EchelonFormMod_gr_67.gif]  the theorem guarantees a unique solution.  

We are really done.

Aside.  The following Mathematica steps will also solve the problem automatically.  
It starts with the equations, creates the matrices, and ends up with the vector form of the solution.  
This is just for fun !

[Graphics:../Images/EchelonFormMod_gr_68.gif]


[Graphics:../Images/EchelonFormMod_gr_69.gif]

 

[Graphics:../Images/EchelonFormMod_gr_70.gif]

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(c) John H. Mathews 2004